Please help me! I do not know what to do.
500.0 g ethyl alcohol at 25°C (C = 2.46 J/g x °C) absorbs 74.0 J of heat energy. What is the final temperature?
What is the specific heat constant if 621 g of Metal A raises the temperature of 500.0 g of water by 3.20°C and the temperature of the metal decreases by 42.5°C.
heat=mc deltaT
deltaT=74.0/(500*2.46)= ?
now final temp, equals initial temp 25C + deltatemp
For part B
heat lost by metal + heat gained by water = 0
[mass metal x sp.h. metal x 43.5] + [mass H2O x sp.h water x 3.20] = 0
Solve for sp.h. (specific heat) metal
To find the final temperature in the first question, you can use the formula for heat transfer:
Q = m * C * ΔT
Where:
Q is the heat energy gained or lost by the substance
m is the mass of the substance
C is the specific heat capacity of the substance
ΔT is the change in temperature
In this case, the formula becomes:
74.0 J = 500.0 g * C * (Tf - 25°C)
To solve for the final temperature (Tf), we need to rearrange the formula:
Tf - 25°C = 74.0 J / (500.0 g * C)
Tf - 25°C = 0.148 J/(g°C) / C
Now, to find the specific heat constant in the second question, you can again use the formula:
Q = m * C * ΔT
In this case, the formula becomes:
C = Q / (m * ΔT)
Substituting the given values:
C = 621 g * C * (-42.5°C - 3.20°C)
Simplifying the equation:
C = -621 g * C * (-45.70 °C)
To get the specific heat constant C, we need to rearrange the formula:
C = -621 g * C / (-45.70 °C)